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moongazer

  • 2 years ago

If tan (theta) =5/4 and cos (theta) < 0 . Find the other trig. functions. Could someone check my answers?

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  1. moongazer
    • 2 years ago
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    am I right that theta is in quadrant 3 ?

  2. Silent_Sorrow
    • 2 years ago
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    Yup, so Sin (theta) < 0 as well

  3. moongazer
    • 2 years ago
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    Could you show your answers so that I can check my answers?

  4. waterineyes
    • 2 years ago
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    No.. you should give your answers so that we can check.. Go ahead..

  5. moongazer
    • 2 years ago
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    ok

  6. moongazer
    • 2 years ago
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    \[\sin \theta =(-5\sqrt{41})/41\] \[\cos \theta = (-4\sqrt{41})/41\] \[\tan \theta = 5/4\] \[\csc \theta = -\sqrt{41}/5\] \[\sec \theta =-\sqrt{41}/4\] \[\cot \theta = 4/5\]

  7. moongazer
    • 2 years ago
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    Am I correct?

  8. Silent_Sorrow
    • 2 years ago
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    |dw:1341743660142:dw| Just remember "All Science Teachers Crazy" so in the first quad all trig functions are positive, in the 2nd all Sine functions are positive, in the 3rd all Tangent functions are positive so in the 4th all Cosine functions are positive.

  9. moongazer
    • 2 years ago
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    Are all my answers correct?

  10. moongazer
    • 2 years ago
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    Could someone check it?

  11. moongazer
    • 2 years ago
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    @mathslover Are my answers correct?

  12. mathslover
    • 2 years ago
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    when cos theta is negative then theta lies in 2nd quadrant ..

  13. moongazer
    • 2 years ago
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    @mathslover why?

  14. mathslover
    • 2 years ago
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    please wait

  15. moongazer
    • 2 years ago
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    ok thanks :)

  16. mathslover
    • 2 years ago
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    Remember this :Quadrant 1: Cosine is positive and Sine is positive. (x > 0 ; y > 0) just as ( cos(theta) > 0 ; sin(theta) > 0) Quadrant 2: Cosine is negative and Sine is positive. (x < 0; y > 0) just as ( cos(theta) < 0 ; sin(theta) > 0) Quadrant 3: Cosine is negative and Sine is negative. (x < 0 ; y < 0) just as ( cos(theta) < 0 ; sin(theta)< 0) Quadrant 4: Cosine is positive and Sine is negative. ( x > 0 ; y < 0 ) just as ( cos(theta) > 0 ; sin(theta) < 0)

  17. mathslover
    • 2 years ago
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    @moongazer

  18. moongazer
    • 2 years ago
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    I am here

  19. mathslover
    • 2 years ago
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    So did u get that quadrant 2 : cos theta is negative ?

  20. moongazer
    • 2 years ago
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    i'm still reading it :)

  21. mathslover
    • 2 years ago
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    Take ur time

  22. moongazer
    • 2 years ago
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    But in quadrant 3 cos theta is also negative so why I can't choose it as the location of theta?

  23. moongazer
    • 2 years ago
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    isn't it that tan theta is positive in quadrant 3 and according to the given tan theta is positive. So choosing that theta is in quadrant 3 is better?

  24. mathslover
    • 2 years ago
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    oh yes u r also right but i am not perfect in this topic may be @lalaly do it for u

  25. moongazer
    • 2 years ago
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    @lalaly My answers are correct right?

  26. lalaly
    • 2 years ago
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    u can describe this as a right triangle \[\tan (\theta)=\frac{opposite}{adjacent}=\frac{5}{4}\]so from this u know that opposite=5 and adjacent =4 |dw:1341746704835:dw|

  27. lalaly
    • 2 years ago
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    now use those values to find the ratios for the other trig functions

  28. moongazer
    • 2 years ago
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    But isn't it I can also view as|dw:1341747710304:dw|

  29. moongazer
    • 2 years ago
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    Are my answers right or wrong?

  30. ash2326
    • 2 years ago
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    @moongazer you are correct :D

  31. moongazer
    • 2 years ago
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    @ash2326 Thanks for that!

  32. ash2326
    • 2 years ago
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    you wanna see, the pic I made to check this?

  33. moongazer
    • 2 years ago
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    ok, let me see it :)

  34. ash2326
    • 2 years ago
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    here it is

    1 Attachment
  35. moongazer
    • 2 years ago
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    That is also what I am thinking :)

  36. ash2326
    • 2 years ago
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    Yay:D

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